Exploration 2: Products

November 9, 2023.

Let’s study subgroups. Let’s say $H$ and $K$ are subgroups of $G$ . How much of $G$ is captured by these two subgroups? In other words, how much of $G$ can you reconstruct from $H$ and $K$ alone?

One obvious answer is to combine the two subgroups. We can take the group product $HK$ , often just called the product, as all elements $hk$ for all $h\in H,k\in K$ . It is not necessarily a group. $HK=\{hk\mid h\in H, k\in K\}$ Of course this only works because $H$ and $K$ are subgroups of the same group, which gives us the definition of a product between elements of $H$ and $K$ . If $H$ and $K$ were arbitrary groups, we would have to come up with a new definition of product between them. More on that later.

$HK$ contains $H$ , and it also contains $K$ . To see this, remember that the identity $e\in G$ exists in both $H$ and $K$ , so $HK$ includes the elements $he=h$ for all $h\in H$ and $ek=k$ for all $k\in K$ . Thinking about group order, this means $HK$ is at least as large as $H$ and $K$ . And $HK$ is at most as large as $G$ , since elements of $HK$ come from $G$ and therefore are contained in $G$ . So: $|H|,|K|\le |HK|\le |G|$

Verify that when $K\subseteq H$ , $HK=H$ , and when $H\subseteq K$ , $HK=K$ . The question is, when is $HK=G$ ? In other words, when do $H$ and $K$ capture all the information about $G$ ? For this to be true, every element of $G$ must be expressible in the form $hk$ for some $h\in H,k\in K$ . Since there are $|H|$ choices for $h$ and $|K|$ choices for $k$ , there are $|H||K|$ choices for $hk$ , which must be at least $|G|$ in order to represent all of $G$ . So, firstly, in order for $HK=G$ to be true, we must have at the very least: $|H|,|K|\le |HK|\le |G|\le |H||K|$

Secondly, we would like to know how many of these $hk$ represent unique elements of $HK$ . To do this, we need to explore when two $hk$ are equal, and we can take their equivalence classes as the unique elements of $HK$ .

Theorem: For finite subgroups

H

and

K

|HK|=|H||K|/|H\cap K|

Consider all pairs $(h,k)$ . There are $|H||K|$ of them, each corresponding to a product $h\cdot k\in HK$ . But not all products are necessarily distinct: $h\cdot k=hx\cdot x^{-1}k$ holds for every element $x\in H\cap K$ .
This means the equivalence class of pairs $(h,k)$ whose product is equal to $h\cdot k$ consists of $|H\cap K|$ pairs. Dividing the number of pairs $|H||K|$ by the size of each equivalence class $|H\cap K|$ gives the number of distinct products $|HK|$ .

Corollary: If $H$ and $K$ have trivial intersection, $|HK|=|H||K|$ .

So in order to have $HK=G$ , we can consider $|HK|=|H||K|=|G|$ when $H,K$ have trivial intersection.

Theorem:

HK=G

exactly when

G

has subgroups

H,K

where

|H\cap K|=\{e\}

and

|H||K|\ge |G|

By the previous theorem $|HK|=|H||K|/|H\cap K|$ , when $|H\cap K|=\{e\}$ we get $|HK|=|H||K|$ . The assumption $|H||K|\ge|G|$ then gives $|HK|\ge|G|$ . But we always have $HK\subseteq G$ and therefore $|HK|\le |G|$ , thus $|HK|=|G|$ . Since all the $|HK|=|G|$ distinct elements in $HK$ can only come from $G$ , $HK=G$ .

Conversely, if $HK=G$ , then $|HK|=|G|$ and every element of $G$ is expressible by a unique pair $(h,k)$ for $h\in H$ and $k\in K$ , implying that $|H\cap K|=\{e\}$ using the same argument in the above theorem. This means $|HK|=|H||K|$ , which together with $|HK|=|G|$ implies $|H||K|=|G|$ , which is enough to show $|H||K|\ge |G|$ as well.

Proving that $|H||K|\ge |G|$ requires knowing $|H||K|$ and $|G|$ , which isn’t often the case, so let’s go in a different direction.

Whenever $H\cap K$ is trivial, we can make an argument about commutativity between elements $h\in H$ and $k\in K$ . In particular, if we can prove that their commutator $hkh^{-1}k^{-1}$ is in both $H$ and $K$ , then we know $hkh^{-1}k^{-1}=e$ and therefore all $h\in H$ commute with all $k\in K$ . When is this true?

Note that $hkh^{-1}k^{-1}\in H$ implies $kh^{-1}k^{-1}\in h^{-1}H=H$ , which exactly means that $H$ is invariant under conjugation by elements of $K$ . By a similar argument, $hkh^{-1}k^{-1}\in K$ implies $hkh^{-1}\in Kk=K$ which exactly states that $K$ is invariant under conjugation by elements of $H$ . We can make both statements true if we assert that $H$ and $K$ are invariant under conjugation by elements in $G$ — in other words, $H$ and $K$ are normal subgroups of $G$ .

Then the logic goes like this. If $H$ and $K$ are normal subgroups of $G$ , then $hkh^{-1}k^{-1}$ is in both $H$ and $K$ , implying that $hkh^{-1}k^{-1}\in H\cap K$ . If $H\cap K$ is trivial, then $hkh^{-1}k^{-1}=e$ meaning that every element of $H$ commutes with every element of $K$ . Then:

Theorem: When

G

has normal subgroups

H,K

with trivial intersection

|H\cap K|=\{e\}

, then

H,K

commute.

We can prove two facts about the commutator of arbitrary $h\in H$ and $k\in K$ . - Since $H$ is normal, $kh^{-1}k^{-1}\in H$ , so $h(kh^{-1}k^{-1})\in H$ . - Since $K$ is normal, $hkh^{-1}\in K$ , so $(hkh^{-1})k^{-1}\in K$ . Therefore the commutator is in both $H$ and $K$ . Since their intersection is trivial, the commutator must be the identity every time, implying that all elements of $H$ commute with all elements of $K$ .

In this section, we try a different approach to constructing a group from given subgroups.

If $H$ and $K$ are both normal with trivial intersection, their group product $HK$ is sometimes called the internal direct product.

In particular, if $|H\cap K|=\{e\}$ , then every single $hk$ is a unique element of $HK$ , and together with $|H||K|\ge |G|$ we get $|HK|=|H||K|=|G|$ . That means every pair $(h,k)$ for $h\in H$ and $k\in K$ can be identified with a unique element of $G$ . In fact, every element of the group $G$ can be written as a pair $(h,k)$ . In general, if two groups are renamings of each other, then they are isomorphic, and the renaming is an isomorphism. Here, we say that $G$ and $S=\{(h,k)\mid h\in H,k\in K\}$ are isomorphic (denoted $G\iso S$ ), and that the bijective renaming $g\mapsto (h,k)$ is an isomorphism.

This method of taking all pairs $(g,h)$ for $g\in G$ and $h\in H$ works for arbitary groups $G,H$ , and is known as the direct product of groups, and is denoted $G\times H$ . It is always a group because product and inverse can be defined componentwise, for example $(g_1,h_1)\cdot(g_2,h_2)=(g_1g_2,h_1h_2)$ .Sometimes the direct product is also called the direct sum, denoted $G\oplus H$ , but that is specifically for abelian groups whose group operation is denoted as addition ( $+$ ).

Theorem:

H\times K\iso G

exactly when

G

has subgroups

H,K

where

|H\cap K|=\{e\}

|H||K|=|G|

, and

hk=kh

for all

h\in H,k\in K

Note that these three conditions are exactly the behavior of $H\times\{e\},\{e\}\times K$ in $H\times K$ :

$(h,e)=(e,k)$ implies $h=e,k=e$ , thus their intersection is trivial
$\<(h,e),(e,k)\>=G$ implies $HK=G$ and therefore $|H||K|=|G|$
$(h,e)(e,k)=(e,k)(h,e)$ is just $hk=kh$ .

That proves the forward direction. To show the reverse direction, we can construct this isomorphism directly with the given conditions.

A common misconception is that the direct product is kind of like the inverse of a group quotient. However, $H\times G/H\not\iso G$ in general.

Theorem:

G/H\times H\not\iso G

Take the example $G=C_4=\<g\>$ .
Let $H=\<g^2\>\iso C_2$ . $H\lhd G$ since $H$ is cyclic, therefore abelian, therefore normal.
Then $G/H$ sends $g^2$ to $e$ , and the result is isomorphic to $C_2$ .
But $C_2\times C_2\not\iso C_4$ , since $C_4$ has an order $4$ element while $C_2\times C_2$ doesn’t.

Here are some other interesting theorems about direct product.

Theorem:

G\times H\iso H\times G

There’s a simple bijection between each $(g,h)\in G\times H$ and $(h,g)\in H\times G$ (swap the elements), therefore the two groups are isomorphic.

Lemma: The order of a direct product

|G_1\times G_2\times\ldots\times G_n|

is equal to

\prod_{i=1}^n|G_i|

Proving this for $|G_1\times G_2|$ is enough to prove the theorem by induction. Since every element of $G_1\times G_2$ chooses an element $g_1\in G_1$ and an element $g_2\in G_2$ , there are $|G_1|$ possible choices for $g_1$ and $|G_2|$ possible choices for $g_2$ , so there are $|G_1||G_2|$ elements in $G_1\times G_2$ , as required.

Chinese Remainder Theorem:

C_m\times C_n\iso C_{mn}

when

\gcd(m,n)=1

(i.e.

m,n

are coprime).

Given $C_m=\<a\>$ and $C_n=\<b\>$ , we want to show $C_{mn}=\<(a,b)\>$ .
$C_m\times C_n$ is generated by $(a,b)$ if every one of its elements $(a^i,b^j)\in C_m\times C_n$ can be written as $(a,b)^k$ for some $k$ .
This amounts to solving the system $\begin{aligned} k&\equiv i\mod m\\ k&\equiv j\mod n\\ \end{aligned}$ for $k\mod mn$ .
Chinese Remainder Theorem: In the above equations, $k$ has only one solution mod $mn$ .
- This is a number-theoretic proof.
- Bezout’s identity: if $\gcd(m,n)=1$ , then there exist integers $c,d$ such that $cm+dn=1$ .
- Then $k$ has a solution: $\begin{aligned} k&=idn+jcm\\ k&=i(1-cm)+jcm&k&=idn+j(1-dn)&\\ k&=i-icm+jcm&k&=idn+j-jdn&\\ k&=i+(j-i)cm&k&=(i-j)dn+j&\\ k&\equiv i\mod m&k&\equiv j\mod n&\\ \end{aligned}$
- To prove the solution unique mod $mn$ , consider another solution $k'$ where $k'\equiv i\mod m$ and $k'\equiv j\mod n$ .
- But then $k'-k\equiv 0\mod m$ and $k'-k\equiv 0\mod n$ , therefore $k'-k$ must be a multiple of both $m$ and $n$ .
- Since $\gcd(m,n)=1$ , $k'-k$ must be a multiple of $mn$ , therefore $k'-k\equiv 0\mod mn$ and $k'\equiv k\mod mn$ .
Then there is exactly one solution for $k$ for every pair $(i,j)$ . So we can write a bijection between each $(a^i,b^j)\in C_m\times C_n$ and the corresponding $(a,b)^k\in C_{mn}$ , which means the two groups are isomorphic.

The direct product discussed above is known as the external direct product $G\times H$ . If both groups are actually subgroups $H,K$ of some encompassing group $G$ , then we can define the internal direct product $HK$ , whose elements are all products $hk$ for $h\in H,k\in K$ . Unlike the external direct product, the internal direct product is restricted to working with elements of the encompassing group. Let’s explore the case when $G=HK$ :

Lemma: When

G=HK

for normal subgroups

H,K\lhd G

, then

H\cap K=\{e\}

iff every

g\in G

factors uniquely into the product of some

h\in H,k\in K

( $\to$ $\to$ )
- $G=HK$ implies that every $g\in G$ can be written as $hk$ for some $h\in H,k\in K$ .
- To prove that this factoring is unique given $H\cap K=\{e\}$ , let $g=h_1k_1=h_2k_2$ . Then: $\begin{aligned} g=h_1k_1&=h_2k_2\\ h_2^{-1}h_1&=k_2k_1^{-1}\\ h_2^{-1}h_1&=k_2k_1^{-1}\in H\cap K\text{ (since }h_2^{-1}h_1\in H\text{ and }k_2k_1^{-1}\in K\text{)}\\ h_2^{-1}h_1&=k_2k_1^{-1}=e\\ h_1=h_2&\text{ and }k_2=k_1 \end{aligned}$ shows that any two such factorings are the same.
( $\from$ $\leftarrow$ )
- Given that $g=hk$ is a unique factoring for all $g\in G$ , let $x$ be an arbitrary element in $H\cap K$ .
- Then $x$ has two factorings $x=x\cdot e=e\cdot x$ , which must be the same due to the unique factoring in $G$ . Then $x=e$ , implying the only element in the intersection $H\cap K$ is $e$ .

Internal Direct Product Theorem: For a group

G

with normal subgroups

H,K\lhd G

whose intersection is trivial and

G=HK

, if

|G|=|H||K|

, then

HK\iso H\times K

By the previous theorem, $H\cap K=\{e\}$ and $G=HK$ implies any element $g\in G$ can be factored uniquely into $hk$ where $h\in H,k\in K$ . This unique factorization essentially maps each $g=hk\in G$ to a distinct pair $(h,k)$ . When $|G|=|H||K|$ , this describes an isomorphism $G=HK\iso H\times K$ .

In this section, we study the semidirect product.

For arbitrary subgroups $H,K$ of $G$ , recall that $HK$ is not necessarily a subgroup.

Theorem: The product of two subgroups of

G

is a not in general a subgroup of

G

The typical counterexample is $\<h\>\<k\>$ with no relations between $h$ and $k$ (each string consisting of copies of $h,k$ represents a unique element.) It cannot be closed under inverses since $(hk)^{-1}=k^{-1}h^{-1}$ , and $k^{-1}h^{-1}$ is distinct from any $h^ak^b\in\<h\>\<k\>$ .

Theorem: The product

HK

of two subgroups

H,K\le G

is a subgroup of

G

if either

H

K

is normal.

Both subgroups contain $e$ , so their product contains $e$ .
Both subgroups $H,K$ are closed under product. To see this, take an arbitrary element $(h_1k_1)(h_2k_2)\in HK$ . Since either $H$ or $K$ is normal, $k_1h_2$ is equal to some $h_2k_1'$ (if $K$ is normal) or some $h_2'k_1$ (if $H$ is normal). Either way, we obtain some $(h_1h_2')(k_1'k_2)$ which is in $HK$ , so $HK$ is closed under product.
Both subgroups are closed under inverses. To see this, take an arbitrary element $hk\in HK$ , whose inverse is $(hk)^{-1}=k^{-1}h^{-1}$ . Since either $H$ or $K$ is normal, $k^{-1}h^{-1}$ is equal to some $h^{-1}k'$ (if $K$ is normal) or some $h'k^{-1}$ (if $H$ is normal). Either way, we obtain some $h'k'$ which is in $HK$ , so $HK$ is closed under inverse.
Since the product $HK$ contains identity, and is closed under product and inverses, it is also a subgroup of $G$ .

Essentially, the reason $hkh'k'$ is in $HK$ if say $H$ is normal is because the normality of $H$ gives us, for each $k\in K$ , a bijection $\varphi_k:H\to H$ defined by $h''\mapsto kh'k^{-1}$ . This is the same as $kh'\mapsto h''k$ , which lets us turn every $hkh'k'$ into $hh''kk'\in HK$ , making $HK$ closed under product as required.

Note that the key part that makes $HK$ a subgroup is not the normality of $H$ , but the bijection $\varphi_k:H\to H$ it defines for every $k\in K$ . This bijection lets us turn every instance of $kh'$ into $\varphi_k(h')k$ for some element $\varphi_k(h')\in H$ . Above, the bijection was simply conjugation by $k$ , but $HK$ is a subgroup as long as we can define $h'\mapsto\varphi_k(h')$ . To see this clearly, suppose you have $hkh'k'$ again, and we want to prove it is in $HK$ . Then since every instance of $kh'$ is equal to $\varphi_k(h')k$ for some $\varphi_k(h')\in H$ , we get $h\varphi_k(h')kk'$ and we are done since $h\varphi_k(h')\in H$ and $kk'\in H$ .

Theorem: The product

HK

of two subgroups

H,K\le G

is a subgroup of

G

if each

k\in K

defines a bijection

\varphi_k:H\to H

such that

khk^{-1}\in H

. khk^{-1} k(kh^{-1}){-1}

The proof of this is similar to the previous one.

Both subgroups contain $e$ , so their product contains $e$ .
Both subgroups $H,K$ are closed under product. Then take an arbitrary element $(h_1k_1)(h_2k_2)\in HK$ . This element is equivalent to
Since either $H$ or $K$ is normal, $k_1h_2$ is equal to some $h_2k_1'$ (if $K$ is normal) or some $h_2'k_1$ (if $H$ is normal). Either way, we obtain some $(h_1h_2')(k_1'k_2)$ which is in $HK$ , so $HK$ is closed under product.
Both subgroups are closed under inverses. Then take an arbitrary element $hk\in HK$ , whose inverse is $(hk)^{-1}=k^{-1}h^{-1}$ . Since either $H$ or $K$ is normal, $k^{-1}h^{-1}$ is equal to some $h^{-1}k'$ (if $K$ is normal) or some $h'k^{-1}$ (if $H$ is normal). Either way, we obtain some $h'k'$ which is in $HK$ , so $HK$ is closed under inverse.
Since the product $HK$ contains identity, and is closed under product and inverses, it is also a subgroup of $G$ .

Just like with the direct product, we can generalize this to $H,K$ that are not subgroups of the same group. Define the semidirect product $H\rtimes_\varphi K$ as the same thing as the direct product $H$ , except the product $(h,k)\cdot(h',k')$ is not $(hh',kk')$ but instead $(h\varphi_k(h'),kk')$ . The idea for the notation $\rtimes$ is that it contains a small $\lhd$ that indicates that $H$ acts ‘normally’ in the sense that $\varphi_k(h')$ exists for each $k\in K$ . Indeed:

Theorem:

H

is a normal subgroup of the semidirect product

H\rtimes\varphi K

This means that

This lets us define a classification theorem for semidirect products. IF $G=HK$ for subgroups $H,K\le G$ where $H\lhd G$ , and $H\cap K$ is trivial then $G$ is isomorphic to $H\rtimes_\varphi K$ where $\varphi$ is conjugation.

TODO wreath product?

zappa-step product?

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